For each of the following solutions, the mass of solute...

For each of the following solutions, the mass of solute taken is indicated, along with the total volume of solution prepared. Calculate the normality of each solution. $\begin{array}{ll}{\text { a. } 113 \mathrm{g} \mathrm{NaOH} ;} & {10.2 \mathrm{mL}} \\ {\text { b. } 12.5 \mathrm{mg} \mathrm{Ca}(\mathrm{OH})_{2} ;} & {100 . \mathrm{mL}} \\ {\text { c. } 12.4 \mathrm{g} \mathrm{H}_{2} \mathrm{SO}_{4} ;} & {155 \mathrm{mL}}\end{array}$

For each of the following solutions, the mass of solute taken is indicated, along with the total volume of solution prepared. Calculate the normality of each solution.
$\begin{array}{ll}{\text { a. } 113 \mathrm{g} \mathrm{NaOH} ;} & {10.2 \mathrm{mL}} \\ {\text { b. } 12.5 \mathrm{mg} \mathrm{Ca}(\mathrm{OH})_{2} ;} & {100 . \mathrm{mL}} \\ {\text { c. } 12.4 \mathrm{g} \mathrm{H}_{2} \mathrm{SO}_{4} ;} & {155 \mathrm{mL}}\end{array}$

Key Concepts

Normality

Normality is a concentration unit that expresses the number of equivalents of a solute per liter of solution. It is particularly useful in acid?base reactions and redox reactions where the reactive capacity of the solute, rather than its molar concentration, is important. In these calculations, determining the number of equivalents requires understanding the reaction stoichiometry of the solute in the solution.

Equivalent Weight

Equivalent weight is defined as the mass of a substance that supplies or reacts with one mole of reactive species (such as H+ in acids or OH– in bases). In acid-base chemistry, this concept is vital because many compounds (for example, Ca(OH)2 or H2SO4) can furnish or accept more than one hydrogen ion or hydroxide ion. Calculating equivalent weight allows conversion from mass to equivalents, which is a key step in determining normality.

Molar Mass

Molar mass is the mass of one mole of a chemical compound and is typically expressed in grams per mole. It serves as a bridge between the mass of the solute and the number of moles, which can then be further converted to equivalents. Precise knowledge of the molar mass of the substances involved is essential for accurate conversion from mass to moles in these concentration calculations.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. In the context of calculating normality, stoichiometry is used to determine how many equivalents a molecule provides based on its chemical structure. For instance, a molecule that dissociates to offer two reactive ions will contribute twice the number of equivalents compared to a molecule that provides only one. This understanding enables the correct scaling from molarity to normality.

Volume and Unit Conversions

Volume and unit conversions involve converting units of volume (for instance, converting milliliters to liters) and ensuring consistency in the units used during calculations. Proper unit conversion is critical in concentration calculations to accurately determine normality, as it allows the computed equivalents to be correctly related to the volume of solution in liters.

Transcript

00:01 This is question 44 from chapter 15 and asking to calculate the normality of each solution.

00:09 So for a, we have 0 .113 grams sodium hydroxide and 10 .2 milliliters.

00:17 So the first thing to do is actually solve for molds for this for sodium hydroxide.

00:21 I'm going to do that.

00:22 The first thing we do is find the molar mass of sodium hydroxide and that is 40 grams.

00:32 After that, we have to take the grams of our solute over the molar mass.

00:41 So it's going to be 0 .1 .13 grams divided by 40.

00:54 And this is going to come out to be 2 .8 times 10 to the negative 3.

01:10 Now after we have that, we can solve for the normality since we also have the milliliters, but the middle leaders have to be put in liters.

01:18 So we're going to divide by 1 ,000, and that is going to give us 0 .0 .02 for the leader.

01:28 So now we can put the 2 .83 times 10 to the negative 3 over 0 .01 .1.

01:50 02 liters and that is going to give us an abnormality of 0 .277 in for an a -o -h.

02:13 Okay, so now we can go on to b.

02:17 And for b, we have 12 .5 milligrams of calcium hydroxide and 100 miliolids.

02:24 Liters.

02:26 So the first thing we should do is convert the milligrams to grams.

02:33 So you do that by dividing by 1000...

Please give Ace some feedback